Question

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$

• A. $\frac{1}{5}$
• B. $\frac{1}{10}$
• C. $-\frac{1}{10}$
• D. $-\frac{1}{5}$

Hint:

Whenever you see a term like $(\sqrt{x}-1)$ in a limit involving polynomials, try to factorize the polynomial as a difference of squares or recognize $(x-1)$ as $(\sqrt{x}-1)(\sqrt{x}+1)$.

Answer 1

The Correct Option is C

Step 1: Spot the trouble.
Direct substitution of $x=1$ gives $\tfrac{0}{0}$, so a common factor must cancel.

Step 2: Factor the denominator.
$2x^2+x-3 = (2x+3)(x-1)$. Also $x-1 = (\sqrt{x}-1)(\sqrt{x}+1)$, so the bottom becomes $(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)$.

Step 3: Cancel the matching factor.
The top has $(\sqrt{x}-1)$, so it cancels with the same factor below, leaving \[ \lim_{x\to 1} \frac{2x-3}{(2x+3)(\sqrt{x}+1)}. \]

Step 4: Substitute now.
At $x=1$: $\tfrac{2-3}{(5)(2)} = \tfrac{-1}{10}$. \[ \boxed{-\tfrac{1}{10}} \]